# Prove these formulas for the distance between and a line/plane.

• Jow
In summary, the first formula for the distance between point B and line L is given by the slope of the line. The second formula for the distance between point B and plane P is given by the slope of the perpendicular line.
Jow
I have two problems that are very similar.
Problem 1: Prove that in the case where the line (L) is in ℝ2 and its equation has the general form ax+by=c, the distance from point B=(x$_0{}$,y$_0{}$) to the line d(B,L) is given by the first formula.

Problem 2: Prove that, in general, the distance d(B, P) from the point B=(x$_0{}$,y$_0{}$,z$_0{}$) to the plane whose general equation is ax+by+cz=d is given by the second formula.

2. Formula 1: d(B,L)=$\frac{\left|ax_0{}+by_0{}-c\right|}{\sqrt{a^2+b^2}}$

Formula 2: d(B,L)=$\frac{\left|ax_0{}+by_0{}+cz_0{}-d\right|}{\sqrt{a^2+b^2+c^2}}$

3. I don't even know where to begin with these.

Jow said:
I have two problems that are very similar.
Problem 1: Prove that in the case where the line (L) is in ℝ2 and its equation has the general form ax+by=c, the distance from point B=(x$_0{}$,y$_0{}$) to the line d(B,L) is given by the first formula.

Problem 2: Prove that, in general, the distance d(B, P) from the point B=(x$_0{}$,y$_0{}$,z$_0{}$) to the plane whose general equation is ax+by+cz=d is given by the second formula.

2. Formula 1: d(B,L)=$\frac{\left|ax_0{}+by_0{}-c\right|}{\sqrt{a^2+b^2}}$

Formula 2: d(B,L)=$\frac{\left|ax_0{}+by_0{}+cz_0{}-d\right|}{\sqrt{a^2+b^2+c^2}}$

3. I don't even know where to begin with these.

Find a unit normal vector to the line. Then you can find the distance by taking the dot product of that vector with the difference between (x0,y0) and any point on the line. The same approach will work for the plane.

That makes sense but you say to use the difference between (x0, y0) and any point on the line. How would you do that? Wouldn't the point you would need to use be (x, y). Is this right?

Jow said:
That makes sense but you say to use the difference between (x0, y0) and any point on the line. How would you do that? Wouldn't the point you would need to use be (x, y). Is this right?

No. You want to get a specific point on the line. Put say x=0 into the equation for the line. What's the corresponding point?

It doesn't give an equation for the line. It just says to prove the formula.

Jow said:
It doesn't give an equation for the line. It just says to prove the formula.

It says the line is ax+by=c. If x=0 what is y?

Another way to do this problem is to use the geometric fact that the distance from a point to a line or a plane is always measured along a line perpendicular to the line or plane.

If a line is given by ax+ by= c, what is the slope of the line? What is the slope of a perpendicular line? What is the equation of a line having that slope and going through $(x_0, y_0)$?

If the plane is given by ax+ by+ cz= d, what is a vector perpendicular to that plane? What are parametric equations for a line in the direction of that vector through $(x_0, y_0, z_0)$?

I get it now. Thanks.

## 1. How do you calculate the distance between a point and a line?

The formula for calculating the distance between a point (x0, y0) and a line ax + by + c = 0 is given by:

d = |ax0 + by0 + c| / √(a^2 + b^2)

## 2. What is the formula for finding the distance between a point and a plane?

The formula for calculating the distance between a point (x0, y0, z0) and a plane ax + by + cz + d = 0 is given by:

d = |ax0 + by0 + cz0 + d| / √(a^2 + b^2 + c^2)

## 3. How do you prove the formula for the distance between a point and a line?

To prove the formula for the distance between a point and a line, we can use the concept of perpendicular distance. We can draw a perpendicular line from the point to the given line and use the Pythagorean theorem to find the distance. By solving the resulting equation, we can derive the formula for the distance.

## 4. Can the distance between a point and a line be negative?

No, the distance between a point and a line cannot be negative. The distance is always considered as a positive value, representing the shortest distance between the point and the line.

## 5. Is there a similar formula for the distance between a point and a plane in three-dimensional space?

Yes, the formula for calculating the distance between a point and a plane in three-dimensional space is similar to the two-dimensional case, except it takes into account the third coordinate (z). The formula is given by:

d = |ax0 + by0 + cz0 + d| / √(a^2 + b^2 + c^2)

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